Download Physics 201 Exam 2 Write also your name in the appropriate box of

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Physics 201
Exam 2
Write also your name in the
appropriate box of the scantron
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Multiple choice questions [60 points]
Answer all of the following questions. Read each question carefully. Fill the correct
bubble on your scantron sheet. Each question has exactly one correct answer. All
questions are worth the same amount of points.
1.
A car travels east at constant velocity. The net force on the car is:
A. east
B. west
C. up
D. down
E. zero since the acceleration is 0 and Fnet = ma.
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2.
A circus performer of weight W is walking along a "high wire" as
shown. The tension in the wire is:
Hint: Draw a free body diagram for the circus performer.
A.
B.
C.
D.
approximately W
approximately W/2
much less than W
much more than W
Draw a free body diagram for the performer
r
TPLeftRope
θ
r
TPRightRope
r
WPE
It follows (with T the magnitude of the tension in the rope)
− WPE + 2T sin θ = 0 ⇒ T =
WPE
⇒ T >> WPE since θ is small.
2 sin θ
E. depends on whether he stands on one or two feet
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3.
A 1-N pendulum bob is held at an angle θ from the vertical by a 2-N
horizontal force F as shown. The tension in the string supporting the
pendulum bob (in newtons) is:
A.
B.
C.
D.
E.
cos θ
2/cos θ
sin θ
tan θ
5
Draw a free body diagram for the bob
r
TBS
r
F
r
WBE
2
T BS = W BE + F 2 = 5 N
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4.
A 70 N block and an 35-N block are connected by a string as shown.
If the pulley is massless and the surface is frictionless, the magnitude
of the acceleration of the 70-N block is:
(Recall the similar computation done in lab)
A. 1.6 m/s2
B. 3.3 m/s2
for the 35 N block
r
r
r
W35 E + T35 S = m35 a35 ⇒ T35 S = W35 E − m35 a35 = 35 − m35 a35
And for the 70 N block
r
r
r
r
T70 S + W70 E + N 70T = m70 a 70 ⇒ T70 S = m70 a 70
Use that
T70 S = T35 S and a 70 = a35 = a to get
m70 a = 35 − m35 a ⇒ a =
C. 4.9 m/s2
D. 6.7 m/s2
E. 9.8 m/s2
35
= 3.3m / s 2
(70 + 35) / 9.8
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5.
A horizontal shove of at least 200-N is required to start moving a 800N crate initially at rest on a horizontal floor. The coefficient of static
friction is:
A. 0.25
200N is the minimum force to accelerate the
box. Thus it is equal to the maximum force
of static friction:
200 = µ s 800 ⇒ µ s = 0.25
B.
C.
D.
E.
0.125
0.50
0.75
1.00
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6.
A box with a weight of 50 N rests on a horizontal surface. A person
pulls horizontally on it with a force of 10 N and it does not move. To
start it moving, a second person pulls vertically upward on the box. If
the coefficient of static friction is 0.4, what is the smallest vertical
force for which the box moves?
A. 4 N
B. 10 N
C. 14 N
25 N
Draw a FBD
r
N BT
r
Fv
r
f BT
r
Fh
r
WBE
D.
the normal force acting on the box is
N BT = WBE − Fv
When the box starts moving, the friction
force is equal to Fh
f BT = Fh
But
f BT = µ S N BT = µ S (WBE − Fv )
⇒ Fv = WBE −
E. 35 N
Fh
µS
= 50 −
10
= 25 N
0 .4
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7.
Which of the following five acceleration versus radius graphs is
correct for a particle moving in a circle of radius r with acceleration a
at a constant speed of 10 m/s?
A.
B.
C.
D.
1
2
3
4
E. 5 since a =
v2
r
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8.
A person riding a Ferris wheel is strapped into her seat by a seat belt.
The wheel is spun so that the centripetal acceleration is g. Select the
correct combination of forces that act on her when she is at the top. In
the table, WPE = weight, down; FPB = seat belt force, down; and FPS =
seat force, up.
A. WPE = 0, FPB = mg, FPS = 0
WPE = mg, FPB = 0, FPS = 0
r
Draw a FBD
F
PS
r
WPE
r
FPB
B.
r
r
r
r
WPE + FPS + FPB = ma
along the vertical ( positive is down)
mg − FPS + FPB = mg ⇒ FPB = FPS
B is the only possible answer
C. WPE = 0, FPB = 0, FPS = mg
D. WPE = mg, FPB = mg, FPS = 0
E. WPE = mg, FPB = 0, FPS = mg
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9.
A crate moves to the right on a horizontal surface as a woman pulls
on it with a 10-N force. Rank the situations shown below according to
the work done by the 10-N force, least to greatest. The displacement
is the same for all cases.
3, 2, 1
A.
Use
B.
C.
D.
E.
2,
2,
1,
1,
W = Fd cos θ
andθ1 = 0,θ 2 ≈ 45,θ 3 = 90
1,
3,
3,
2,
3
1
2
3
10. A man pulls a sled along a rough horizontal surface by applying a
r
constant force F at an angle θ above the horizontal. In pulling the
sled a horizontal distance d, the work done by the man is:
A.
B.
C.
D.
E.
Fd
Fd cos θ
Fd sin θ
Fd/cos θ
Can’t tell without knowing the coefficient of kinetic friction.
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11. A particle is initially at rest on a horizontal frictionless table. It is acted
upon by a constant horizontal force F. Which of the following five
graphs is a correct plot of work W as a function of particle speed v?
Hint: use the work energy theorem
A. I
B. II
C. III
D.
W = K f − Ki 

1
1
IV K f = mv 2  ⇒ W = mv 2
2
2

Ki = 0

E. V
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12. The velocity of a particle moving along the x axis changes from vi to vf.
For which values of vi and vf is the total work done on the particle
negative?
A.
B.
C.
D.
E.
vi = 2m/s, vf = 5m/s
vi = –2m/s, vf = 5m/s
vi = –5m/s, vf = 2m/s
vi = 2m/s, vf = –5m/s
vi = –2m/s, vf = –5m/s

W = K f − Ki 

1
1
2
2
2
K f = mv f  ⇒ W = m(v f − vi ) ⇒ W < 0 if v f < vi
2
2

1
2 
K i = mvi 
2

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PROBLEM [40 points]
Winkin, Blinkin, and Nod are identical triplets, each having a
mass m = 30.0 kg. They clasp arms in a line and go out on to
an ice-covered pond.
Winkin grabs onto a post anchored in the ice with his free
arm, and starts to go around the post. He soon reaches a
constant speed, where he makes one complete revolution in a
time T = 6.28 s.
The distance from the center of each boy to the end of his arm is 0.50 m, so that Nod’s body is 2.5
m from the post and Blinkin’s body is 1.5 m from the post. At the instant shown, the boys are lined
up along the x-axis ( iˆ ), and skating at constant speed in the negative y direction ( − ĵ , out of the
page). The z-axis ( k̂ ) is vertical.
1). [10 pts] If the boys’ arms stay rigid, what is Nod’s velocity as he goes around the
post? Measure his velocity at the center of his body which is 2.5 m from the post.
Give the direction, units and magnitude. Explain.
Nod is moving on a circle at constant speed. Nod makes one revolution = 2πr in T.
2πr
v=
= 2.5m / s
T
At the instant shown the direction of Nod’s velocity is − ĵ (as mentioned above)
2). [10 pts] What is Nod’s acceleration vector at the instant shown (direction, units, and
magnitude). Explain.
The acceleration is toward the center of the circle and has magnitude v2/r (Nod is moving on a
circle at constant speed).
r
v2
a Nod = (−iˆ) = −2.5iˆ m / s 2
r
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r
3). [15 pts]What is the force FBN the force with which Nod pulls on Blinkin? Give the
direction, units and magnitude. Explain.
r
r
According the Newton’s 3rd law FBN = − FNB
Draw a FBD for Nod
r
N NI
r
FNB
r
W NE
r
r
FNB = ma N = −30 × 2.5 × iˆ = −75iˆ N
Thus
r
FBN = 75iˆ N
r
4). [10 pts] What is the force FBW the force with which Winkin pulls on Blinkin? Give the
direction, units and magnitude. Explain.
Draw a FBD for Blinkin
r
FBW
r
N BI
r
FBN
r
W BE
r
The velocity of Blinkin is v B = −1.5 ˆj m / s
r
The acceleration of Blinkin is a B = −1.5iˆ m / s 2
r
r
r
r
Thus FBW + FBN = ma B ⇒ FBW = 30 × (−1.5iˆ) − 75iˆ = −120iˆ N